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Ta có xn luôn dương
Ta có \(2x_n+1=\) \(2\times\dfrac{\left(2+cos\alpha\right)x_n+cos^2\alpha}{\left(2-2cos2\alpha\right)x_n+2-cos2\alpha}+1=\)
\(=\dfrac{6x_n+2cos^2\alpha+2-cos2\alpha}{\left(2-2cos2\alpha\right)x_n+2-cos2\alpha}\)
\(=\dfrac{6x_n+2cos^2\alpha+2sin^2a+1}{\left(2x_n+1\right)\left(1-cos2\alpha\right)+1}\)
\(=\dfrac{3\left(2x_n+1\right)}{2\sin^2\alpha\left(2x_n+1\right)+1}\)
\(\Rightarrow\dfrac{1}{2x_{n+1}+1}=\dfrac{2\sin^2\alpha\left(2x_n+1\right)+1}{3\left(2x_n+1\right)}\)
\(=\dfrac{1}{3}\left(2\sin^2\alpha+\dfrac{1}{2x_n+1}\right)\)
\(\Rightarrow\dfrac{1}{2x_{n+1}+1}-\sin^2\alpha=\dfrac{1}{3}\left(\dfrac{1}{2x_n+1}-\sin^2\alpha\right)\)
\(\Rightarrow\dfrac{1}{2x_{n+1}+1}-\sin^2\alpha=\left(\dfrac{1}{3}\right)^n\left(\dfrac{1}{2x_1+1}-\sin^2\alpha\right)\)
\(=\left(\dfrac{1}{3}\right)^n\left(\dfrac{1}{3}-\sin^2\alpha\right)\)
\(\Rightarrow y_n=\sum\limits^{n-1}_{i=0}\left(\dfrac{1}{3}\right)^i\left(\dfrac{1}{3}-\sin^2\alpha\right)+n\sin^2\alpha\)
\(=\dfrac{1-\left(\dfrac{1}{3}\right)^n}{1-\dfrac{1}{3}}\left(\dfrac{1}{3}-\sin^2\alpha\right)+n\sin^2\alpha\)
hãy nhớTừ công thức truy hồi ta có:
\(x_{n+1}>x_n,\forall n=1,2...\)
\(\Rightarrow\)dãy số \(\left(x_n\right)\) là dãy số tăng
giả sử dãy số \(\left(x_n\right)\) là dãy bị chặn trên \(\Rightarrow limx_n=x\)
Với x là nghiệm của pt ta có: \(x=x^2+x\Leftrightarrow x=0< x_1\) (vô lý)
=> dãy số \(\left(x_n\right)\) không bị chặn hay \(limx_n=+\infty\)
Mặt khác: \(\frac{1}{x_{n+1}}=\frac{1}{x_n\left(x_n+1\right)}=\frac{1}{x_n}-\frac{1}{x_n+1}\)
\(\Rightarrow\frac{1}{x_n+1}=\frac{1}{x_n}-\frac{1}{x_n+1}\)
\(\Rightarrow S_n=\frac{1}{x_1}-\frac{1}{x_{n+1}}=2-\frac{1}{x_{n+1}}\)
\(\Rightarrow limS_n=2-lim\frac{1}{x_{n+1}}=2\)
1.
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{x^2-3x-4}}{1-x^2}=\lim\limits_{x\to (-1)-}\frac{\sqrt{(x+1)(x-4)}}{(1-x)(1+x)}\)
\(=\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{(x-1)\sqrt{-(x+1)}}=-\infty\) do:
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{x-1}=\frac{-\sqrt{5}}{2}<0\) và \(\lim\limits_{x\to (-1)-}\frac{1}{\sqrt{-(x+1)}}=+\infty\)
2.
\(\lim\limits_{x\to 2+}\left(\frac{1}{x-2}-\frac{x+1}{\sqrt{x+2}-2}\right)=\lim\limits_{x\to 2+}\frac{1-(x+1)(\sqrt{x+2}+2)}{x-2}=-\infty\) do:
\(\lim\limits_{x\to 2+}\frac{1}{x-2}=+\infty\) và \(\lim\limits_{x\to 2+}[1-(x+1)(\sqrt{x+2}+2)]=-11<0\)
\(x_1=a>2;x_{n+1}=x_n^2-2,\forall n=1,2,...\)
mà \(n\rightarrow+\infty\)
\(\Rightarrow a\rightarrow+\infty\Rightarrow x_n\rightarrow+\infty\)
\(\Rightarrow\lim\limits_{n\rightarrow+\infty}\dfrac{1}{x_n}=0\) \(\Rightarrow\lim\limits_{n\rightarrow+\infty}\left(\dfrac{1}{x_nx_{n+1}}\right)=0\)
\(\)\(\Rightarrow\lim\limits_{n\rightarrow+\infty}\left(\dfrac{1}{x_1}+\dfrac{1}{x_1x_2}+\dfrac{1}{x_1x_2x_3}+...+\dfrac{1}{x_1x_2...x_n}\right)=0\)
...